a,x/0,3=y/0,7=z và z-3x=1 b, x-5/3=y-4/4=z-3/5 và x+y+z=36 c,x-1/3=y-2/4=z+7/5 và x+y-z=8 26/08/2021 Bởi Natalia a,x/0,3=y/0,7=z và z-3x=1 b, x-5/3=y-4/4=z-3/5 và x+y+z=36 c,x-1/3=y-2/4=z+7/5 và x+y-z=8
a. $\dfrac{x}{0,3} = \dfrac{y}{0,7} = z \to \dfrac{x}{0,3} = \dfrac{y}{0,7} = \dfrac{z}{1} = \dfrac{3x}{0.9} = \dfrac{z – 3x}{1 – 0,9} = \dfrac{1}{0,1} = 10$ Suy ra: $\dfrac{x}{0,3} = 10 \to x = 0,3.10 = 3$ $\dfrac{y}{0,7} = 10 \to y = 0,7.10 = 7$ $\dfrac{z}{1} = 10 \to z = 1.10 = 10$b. $\dfrac{x – 5}{3} = \dfrac{y – 4}{4} = \dfrac{z – 3}{5} = \dfrac{x – 5 + y – 4 + z – 3}{3 + 4 + 5} = \dfrac{(x + y + z) – (5 + 4 + 3)}{12} = \dfrac{36 – 12}{12} = \dfrac{24}{12} = 2$ Suy ra: $\dfrac{x – 5}{3} = 2 \to x – 5 = 3.2 \to x = 6 + 5 = 11$ $\dfrac{y – 4}{4} = 2 \to y – 4 = 4.2 \to y = 8 + 4 = 12$ $\dfrac{z – 3}{5} = 2 \to z – 3 = 5.2 \to z = 10 + 3 = 13$ c. $\dfrac{x – 1}{3} = \dfrac{y – 2}{4} = \dfrac{z + 7}{5} = \dfrac{x – 1 + y – 2 – z – 7}{3 + 4 – 5} = \dfrac{x + y – z – 1 – 2 – 7}{3 + 4 – 5} = \dfrac{8 – 10}{2} = – 1$ Suy ra: $\dfrac{x – 1}{3} = – 1 \to x – 1 = – 3 \to x = – 2$ $\dfrac{y – 2}{4} = – 1 \to y – 2 = – 4 \to y = – 2$ $\dfrac{z + 7}{5} = – 1 \to z + 7 = – 5 \to z = – 12$ Bình luận
a. $\dfrac{x}{0,3} = \dfrac{y}{0,7} = z \to \dfrac{x}{0,3} = \dfrac{y}{0,7} = \dfrac{z}{1} = \dfrac{3x}{0.9} = \dfrac{z – 3x}{1 – 0,9} = \dfrac{1}{0,1} = 10$
Suy ra:
$\dfrac{x}{0,3} = 10 \to x = 0,3.10 = 3$
$\dfrac{y}{0,7} = 10 \to y = 0,7.10 = 7$
$\dfrac{z}{1} = 10 \to z = 1.10 = 10$
b. $\dfrac{x – 5}{3} = \dfrac{y – 4}{4} = \dfrac{z – 3}{5} = \dfrac{x – 5 + y – 4 + z – 3}{3 + 4 + 5} = \dfrac{(x + y + z) – (5 + 4 + 3)}{12} = \dfrac{36 – 12}{12} = \dfrac{24}{12} = 2$
Suy ra:
$\dfrac{x – 5}{3} = 2 \to x – 5 = 3.2 \to x = 6 + 5 = 11$
$\dfrac{y – 4}{4} = 2 \to y – 4 = 4.2 \to y = 8 + 4 = 12$
$\dfrac{z – 3}{5} = 2 \to z – 3 = 5.2 \to z = 10 + 3 = 13$
c. $\dfrac{x – 1}{3} = \dfrac{y – 2}{4} = \dfrac{z + 7}{5} = \dfrac{x – 1 + y – 2 – z – 7}{3 + 4 – 5} = \dfrac{x + y – z – 1 – 2 – 7}{3 + 4 – 5} = \dfrac{8 – 10}{2} = – 1$
Suy ra:
$\dfrac{x – 1}{3} = – 1 \to x – 1 = – 3 \to x = – 2$
$\dfrac{y – 2}{4} = – 1 \to y – 2 = – 4 \to y = – 2$
$\dfrac{z + 7}{5} = – 1 \to z + 7 = – 5 \to z = – 12$